Question: The grades on a geometry midterm at Covington are normally distributed with $\mu = 83$ and $\sigma = 3.0$. Kevin earned a $78$ on the exam. Find the z-score for Kevin's exam grade. Round to two decimal places.
Answer: A z-score is defined as the number of standard deviations a specific point is away from the mean We can calculate the z-score for Kevin's exam grade by subtracting the mean $(\mu)$ from his grade and then dividing by the standard deviation $(\sigma)$ $ { z = \dfrac{x - {\mu}}{{\sigma}}} $ $ { z = \dfrac{78 - {83}}{{3.0}}} $ ${ z \approx -1.67}$ The z-score is $-1.67$. In other words, Kevin's score was $1.67$ standard deviations below the mean.